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Let $G(x)=\sum_{k\leq x}\frac{\mu(k)}{k}$, where $\mu$ is the Mobius function. From this question and its answer, its mention the Riemann hypothesis is equivalent to $G(x)=O(x^{-\frac{1}{2}+\epsilon})$.

I have never heard of this, the closet thing I have heard is the Riemann hypothesis is equivalent to $M(x)=O(x^{\frac{1}{2}+\epsilon})$ for all $\epsilon>0$, where $M(x)=\sum_{k\leq x}\mu(k)$, the Mertens function. I have tried to prove the above fact but did not succeed. So I would like to ask for reference for the above equivalent form of the Riemann hypothesis(the $G(x)$ one), better to contain a proof of it.

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    $\begingroup$ Abel's summation formula gives $$ \sum\limits_{n \le x} {\frac{{\mu (n)}}{n}} = \frac{{M(x)}}{x} + \int_1^x {\frac{{M(t)}}{{t^2 }}{\rm d}t} . $$ Can you finish from here? $\endgroup$
    – Gary
    Commented Mar 16, 2023 at 11:29
  • $\begingroup$ Can you write an answer? I try to bounded the integral term, but then the integral seems unbounded when $\epsilon>0$, also there are constant term that didn't cancel out. $\endgroup$
    – Ken.Wong
    Commented Mar 16, 2023 at 12:06
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    $\begingroup$ Some explicit bounds can be found at the link below. One such bound is $$|M(t)|<\frac{t}{\log(t)^{11/9}}$$ for $x>1000$. en.wikipedia.org/wiki/Mertens_function#Known_upper_bounds $\endgroup$
    – QC_QAOA
    Commented Mar 16, 2023 at 12:17

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Abel's summation formula gives $$ G(x):=\sum\limits_{n \le x} {\frac{{\mu (n)}}{n}} = \frac{{M(x)}}{x} + \int_1^x {\frac{{M(t)}}{{t^2 }}{\rm d}t} . $$ Assume RH. Then $M(x)=\mathcal{O}(x^{1/2+\varepsilon})$. By the prime number theorem, $$ \sum\limits_{n = 0}^\infty {\frac{{\mu (n)}}{n}} = 0. $$ Thus $$ \sum\limits_{n \le x} {\frac{{\mu (n)}}{n}} = \frac{{M(x)}}{x} - \int_x^{ + \infty } {\frac{{M(t)}}{{t^2 }}{\rm d}t} , $$ where \begin{align*} \frac{{M(x)}}{x} - \int_x^{ + \infty } {\frac{{M(t)}}{{t^2 }}{\rm d}t} & = \mathcal{O}(x^{ - 1/2 + \varepsilon } ) + \mathcal{O}(1)\int_x^{ + \infty } {\frac{{\rm d}t}{{t^{3/2 - \varepsilon } }}} \\ & = \mathcal{O}(x^{ - 1/2 + \varepsilon } ) + \mathcal{O}(x^{ - 1/2 + \varepsilon } ) = \mathcal{O}(x^{ - 1/2 + \varepsilon } ). \end{align*} Now assume $G(x)=\mathcal{O}(x^{ - 1/2 + \varepsilon } )$. Then, by Abel's summation formula, $$ M(x):=\sum\limits_{n \le x} {\mu (n)} = \sum\limits_{n \le x} {\frac{{\mu (n)}}{n}n} = x\,G(x) + \int_1^x {G(t)\,{\rm d}t} . $$ Using our assumption, it follows that $M(x)=\mathcal{O}(x^{1/2+\varepsilon})$, which we know implies RH.

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  • $\begingroup$ At the third line, why do the integration limit changes? $\endgroup$
    – Ken.Wong
    Commented Mar 16, 2023 at 12:32
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    $\begingroup$ Because I added and subtracted $$ \int_1^{ + \infty } {\frac{{M(t)}}{{t^2 }}{\rm d}t} \quad \left( { = \sum\limits_{n = 0}^\infty {\frac{{\mu (n)}}{n}} = 0} \right). $$ $\endgroup$
    – Gary
    Commented Mar 16, 2023 at 12:33
  • $\begingroup$ Thanks. For the fourth line, if $\epsilon>1/2$, isn't the integral diverges? $\endgroup$
    – Ken.Wong
    Commented Mar 16, 2023 at 12:36
  • $\begingroup$ You can assume that $0<\varepsilon<1/2$. Then the case $\varepsilon \ge 1/2$ follows trivially. $\endgroup$
    – Gary
    Commented Mar 16, 2023 at 12:37
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    $\begingroup$ No need of the PNT if you are assuming that $\sum_{n\ge 1}\mu(n)n^{-s}$ converges for $\Re(s) > 1/2$ as it implies that $\sum_{n\ge 1}\mu(n)n^{-1}=\lim_{s\to 1} \sum_{n\ge 1}\mu(n)n^{-s}=0$ :-) $\endgroup$
    – reuns
    Commented Mar 16, 2023 at 13:17

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